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April 15th, 2014, 03:32 AM   #1
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probability question.

1. A bag contains 10 colored balls: 2 red, 3 blue and 5 yellow. Two balls are drawn from the bag at random, one after the other. Given that the two balls drawn were of the same color, what is the probability that if third ball is drawn, it is also of that color?

2. A lot of 25 is to be inspected by means of a two stage sampling plan. A sample of five sample is drawn. If one or more is bad, the lot is rejected. If all are good, a second sample of 10 item is drawn from the remaining 20 items. The lot is then rejected if any item in the second sample is bad, and is accepted if all 10 items are good. What is the probability that the lot is rejected if it contains 2 bad items?

thank you!!!
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April 15th, 2014, 07:10 AM   #2
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Hello, mcm1234!

Quote:
1. A bag contains: 2 red, 3 blue and 5 yellow balls.
Two balls are drawn from the bag at random, one after the other.
Given that the two balls drawn were of the same color,
what is the probability that if a third ball is drawn, it is also of that color?

Just work out the possible scenarios.

Both red
We have: $\displaystyle \,\{0R,\;3B,\;5Y\} \quad\Rightarrow\quad P(\text{3rd R}) \,=\, 0$

Both blue
We have: $\displaystyle \,\{2R,\,1B,\,5Y\} \quad\Rightarrow\quad P(\text{3rd B}) \,=\,\tfrac{1}{8}$

Both yellow
We have: $\displaystyle \,\{2R,\;3B,\;3Y\} \quad\Rightarrow\quad P(\text{3rd Y}) \,=\, \tfrac{3}{8}$


Therefore: $\displaystyle \:P(\text{3 same color}) \:=\:0 + \tfrac{1}{8} + \tfrac{3}{8} \:=\:\tfrac{1}{2}$

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April 15th, 2014, 07:36 AM   #3
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thanks for the reply!
the answer for no.1 supposed to be 33/112 (my teacher said so), but I don't know how to get that answer.
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April 15th, 2014, 01:11 PM   #4
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You can do it this way:

P(2 Red) = (2/10)(1/9) = 2/90

P(2 Blue) = (3/10)(2/9) = 6/90

P (2 Green) = (5/10)(4/9) = 20/90

So, two balls of the same colour will be picked 28/90. Of those 28 events, the breakdown will be above. And, the probability the third is the same colour is:

2/28* 0 + 6/28 * 1/8 + 20/28*3/8 = 6/224 + 60/224 = 66/224 = 33/112
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April 15th, 2014, 02:51 PM   #5
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Hello again, mcm1234!

I just solved it as a Conditional Probability problem
$\displaystyle \quad$and got your teacher's answer.
So my solution (and reasoning) must be wrong . . . *blush*


Quote:
The answer for no. 1 is supposed to be 33/112 (my teacher said so).

Bayes' Theorem: $\displaystyle \;P(A\,|\,B) \;=\;\frac{P(A \wedge B)}{P(B)}$

{2 red, 3 blue, 5 yellow}

Three balls are drawn.
Given that the first two balls are the same color,
find the probability that the third ball is also the same color.

$\displaystyle P(\text{all 3 same color}\,|\,\text{first 2 same color}) \;=\;\frac{P(\text{all 3 same}\,\wedge\,\text{first 2 same})}{P(\text{first 2 same})} $

The numerator is: $\displaystyle \,P(\text{all 3 same color})$

$\displaystyle P(RRR) \;=\;0$

$\displaystyle P(BBB) \;=\;\tfrac{3}{10}\!\cdot\!\tfrac{2}{9}\!\cdot\! \tfrac{1}{8} \;=\;\tfrac{1}{120}$

$\displaystyle P(YYY) \;=\;\tfrac{5}{10}\!\cdot\!\tfrac{4}{9}\!\cdot\! \tfrac{3}{8} \;=\;\tfrac{1}{12}$

$\displaystyle \quad\text{Hence: }\:P(\text{all 3 same}) \;=\;\tfrac{1}{120} + \tfrac{1}{12} \;=\;\tfrac{11}{120}$


The denominator is: $\displaystyle \,P(\text{first 2 same})$

$\displaystyle P(RR) \:=\:\tfrac{2}{10}\!\cdot\!\tfrac{1}{9} \:=\:\tfrac{1}{45}$

$\displaystyle P(BB) \:=\:\tfrac{3}{10}\!\cdot\!\tfrac{2}{9} \:=\:\tfrac{1}{15}$

$\displaystyle P(YY) \:=\:\tfrac{5}{10}\!\cdot\!\tfrac{4}{9} \:=\:\tfrac{2}{9} $

$\displaystyle \quad\text{Hence: }\:P(\text{first 2 same}) \:=\:\tfrac{1}{45} + \tfrac{1}{15} + \tfrac{2}{9} \:=\:\tfrac{14}{45}$


$\displaystyle \text{Therefore: }\:P(\text{all 3 same}\,|\,\text{first 2 same}) \;=\;\frac{\frac{11}{120}}{\frac{14}{45}} \;=\;\frac{33}{112}$

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April 15th, 2014, 04:36 PM   #6
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Thank you so much for the answer!!
But can somebody help me with no. 2?
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April 15th, 2014, 08:42 PM   #7
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Hello again, mcm1234!

Quote:
2. A lot of 25 is to be inspected by means of a two-stage sampling plan.

A sample of five sample is drawn.
If one or more is bad, the lot is rejected.

If all are good, a second sample of 10 item is drawn from the remaining 20 items.
The lot is then rejected if any item in the second sample is bad,
and is accepted if all 10 items are good.
What is the probability that the lot is rejected if it contains 2 bad items?

There are: 2 bad (B) and 23 good (G).

A sample of 5 is drawn: $\displaystyle \;{25\choose5} = 53,\!130$ outcomes.

To get 5 G, there are: $\displaystyle \;{23\choose5} = 33,\!649$ ways.

$\displaystyle \quad P(\text{all G}) \:=\:\frac{33,\!649}{53,\!130} \:=\:\frac{19}{30}$

$\displaystyle \quad P(\text{pass 1st stage}) \;=\;\frac{19}{30}$


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Now we have: $\displaystyle \:2\,B,\;18\,G.$

A sample of 10 is drawn: $\displaystyle {20\choose10} \,=\, 184,\!756$ outcomes.

To get 10 G, there are: $\displaystyle \:{18\choose10} \,=\,43,\!758$ ways.

$\displaystyle P(\text{all G}) \:=\:\frac{43,\!758}{184,\!756} \:=\:\frac{9}{38}$

$\displaystyle \quad P(\text{pass 2nd stage}) \;=\;\frac{9}{38}$


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


$\displaystyle \text{Hence: }\:P(\text{pass both stages}) \;=\;\frac{19}{30}\cdot\frac{9}{38} \;=\;\frac{3}{20}$


$\displaystyle \text{Therefore: }\:P(\text{Reject}) \;\;=\;\;1-\frac{3}{20} \;\;=\;\;\frac{17}{20}$

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